Question: Square $ABCD$ has sides of length 4, and $M$ is the midpoint of $\overline{CD}$. A circle with radius 2 and center $M$ intersects a circle with radius 4 and center $A$ at points $P$ and $D$. What is the distance from $P$ to $\overline{AD}$? Express your answer as a common fraction.

[asy]
pair A,B,C,D,M,P;
D=(0,0);
C=(10,0);
B=(10,10);
A=(0,10);
M=(5,0);
P=(8,4);
dot(M);
dot(P);
draw(A--B--C--D--cycle,linewidth(0.7));
draw((5,5)..D--C..cycle,linewidth(0.7));
draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,S);
label("$P$",P,N);
[/asy]
Answer: We place the points on a coordinate system: $D$ at the origin, $C$ and $A$ on the positive $x$- and $y$-axes, respectively. Then the circle centered at $M$ has equation \[(x-2)^{2} + y^{2} = 4\]and the circle centered at $A$ has equation \[x^{2} + (y-4)^{2} = 16.\]Solving these equations for the coordinates of $P$ gives $x=16/5$ and $y=8/5$, so the answer is $\boxed{16/5}$.

[asy]
unitsize(0.5cm);
pair A,B,C,D,M,R,P,Q;
A=(0,4);
B=(4,4);
C=(4,0);
D=(0,0);
M=(2,0);
R=(3.2,0);
P=(3.2,1.6);
Q=(0,1.6);
draw((-2.3,0)--(4.7,0),Arrow);
draw((0,-2.3)--(0,4.7),Arrow);
for (int i=-2;i<5; ++i) {
draw((-0.2,i)--(0.2,i));
draw((i,-0.2)--(i,0.2));
}
draw((2.83,1.17)..B--A--D..cycle,linewidth(0.7));
draw(A--B--C--D--cycle,linewidth(0.7));
draw((2,2)..C--D..cycle,linewidth(0.7));
draw(R--P--Q,linewidth(0.7));
dot(P);
label("$Q$",Q,W);
label("$A$",A,W);
label("$D$",D,SW);
label("$M$",M,S);
label("$R$",R,S);
label("$C$",C,S);
label("$P$",P,N);
label("$B$",B,E);
label("$x$",(4.7,0),S);
label("$y$",(0,4.7),E);
[/asy]



We also could have solved this problem with a little trigonometry:

Let $\angle MAD = \alpha$. Then \begin{align*}
PQ &= (PA)\sin(\angle PAQ) \\
&= 4\sin(2\alpha) \\
&= 8 \sin\alpha\cos\alpha\\
&= 8\displaystyle\left(\frac{2}{\sqrt{20}}\right)\left(\frac{4}{\sqrt{20}}\displaystyle\right)\\
&=\boxed{\frac{16}{5}}.
\end{align*}